# prove that $S^1$ is smooth submanifold of $\mathbb{R}^2$ using the definition with diffeomorphism

I'm tring to prove that the unit circle $$S^1=\{(x_1,x_2)\in\mathbb{R}^2\text{ such that }x_1^2+x_2^2=1\}$$ is an embedded submanifold of $$\mathbb{R}^2$$ using the following Characterization:

A nonempty subset $$M \subset\mathbb{R}^n$$ is an m-manifold iff:

For every $$p\in M$$, there are two open sets $$O,W\subset\mathbb{R}^n$$ with $$0_n\in O$$ and $$p ∈ M ∩ W$$, and a smooth diffeomorphism $$ϕ: O → W$$, such that $$ϕ(0_n) = p$$ and $$ϕ(O ∩ (\mathbb{R}^m × {0_{n−m}})) = M ∩ W$$.

My attempt:

lets fix $$a\in S^1$$, there exists a unique $$\theta_0\in[0,2\pi)$$ such that $$a=(cos(\theta_0),sin(\theta_0))$$

and let $$\phi$$ be the diffeomorphism $$\begin{array}{cccc} \phi : & (0,\infty)\times(\theta_0-\pi,\theta_0+\pi) & \longrightarrow & \mathbb{R}^2\backslash D_{\theta_0+\pi}\\ ~~ & (r,\theta) & \mapsto &(rcos(\theta),rsin(\theta)) \end{array}$$

where $$D_{\theta_0+\pi}$$ is the half line at the origin with polar angle $$\theta_0$$

From here i dont know how to proceed. Should i modify this map or use as it is? and who are the two open sets of the charactrization?

The way you constructed your $$\phi$$ is almost good. Intuitively, the characterization you presented requires $$\phi$$ to be a diffeomorphism between 2 open sets $$W$$ - the ambient space around $$\mathbb S^1$$, and $$O$$ - a deformed version of $$W$$ such that $$\mathbb S^1 \cap W$$ is "squished and flattened" into $$\mathbb R \times \left \{ 0 \right \}$$ (by $$\phi^{-1}$$). You could think of $$\phi|_{O\cap (\mathbb R \times \left \{ 0 \right \})}$$ as a parameterization of the circle using $$\theta$$.
Your $$\phi$$ is a good candidate besides the fact there needs to be some alignment so that $$0$$ is mapped to $$a$$: substituting $$\theta+\theta_0$$ into the trig functions instead of $$\theta$$, and using a radius $$1+r$$ instead of $$r$$. This way we get $$W=\mathbb R^2 \setminus D_{\theta_0+\pi}$$ as you said but $$O=(-1,\infty)\times(-\pi,\pi)$$ From the naturality of polar coordinates it's pretty easy to convince $$\phi$$ is bijective, two points can't have the same radius and angle, and every point has a radius and an angle. $$\phi$$ is also smooth due to its components being elementary functions. Hence, it is a diffeomorphism.
$$\phi(O\cap (\mathbb R \times \left \{ 0 \right \}))\subseteq \mathbb S^1 \cap W$$ since any point of the form $$(0,\theta)$$ is mapped to the unit circle at an angle which isn't $$\theta_0+\pi$$, and $$\mathbb S^1 \cap W \subseteq\phi(O\cap (\mathbb R \times \left \{ 0 \right \}))$$ using the reverse argument.
• thank you sir you're a genius So we chose our diffeomorphism to be \begin{array}{cccc} \phi : & (-1,\infty)\times(-\pi,+\pi) & \longrightarrow & \mathbb{R}^2\backslash D_{\theta_0+\pi}\\ ~~ & (r,\theta) & \mapsto &((1+r)cos(\theta+\theta_0),(1+r)sin(\theta+\theta_0)) \end{array} $O=(-1,\infty)\times(-\pi,+\pi)$ and $w=\mathbb{R}^2\backslash D_{\theta_0+\pi}$ – Donnie Darko Oct 23 '20 at 18:28